Biased Random Number Generator

Question

I am looking for a random number generator that can be biased. For instance, say I want a random number between 1-5, with the probability being:

1: Comes up 20% of t

Answer 1

Kenny gave an appropriate answer tailored to your particular frequency distribution.

The more general answer works with a CDF - Cumulative Distribution Function - for the data, and uses a uniform random number to pick a value within the distribution.

Answer 2

For your problem, just pick a random element from this list uniformly:

[1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5]

In general, check this answer: Weighted random numbers

---

In TR1 and C++0x, there is <random> header which contains the discrete_distribution class to generate such numbers, among others.

You may also want to check out GSL which contains much more random distributions (and random number generators) than the standard <random> library. (But note that GSL uses GPLv3.)

Answer 3

Throw a random real number x in [0,1], if 0< x<0.2 return 1, if 0.2<x <0.3 return 2, etc.

See here for the general problem.

Answer 4

Best way's probably to just take the normal unbiased random generator then return based on the interval its value falls into.

Just an if statement that gives 1 for 0:0.2, 2 for 0.2:0.3, 3 for 0.3:0.7, 4 for 0.7:0.95 and 5 for 0.95:1. Best to make either the lower or upper limit of the interval inclusive and the other exclusive.

int biasedRandom(){
double i = randomNumber();
if(i<= 0.2){return 1;}
else if(i <= 0.3){return 2;}
else if(i <= 0.7){return 3;}
else if(i <= 0.95){return 4;}
else{return 5;}
}

Something like that.

Answer 5

Coming late to the party on this one. Here is the C++0x answer:

#include <iostream>
#include <random>
#include <iterator>

int main()
{
    // Set up distribution
    double interval[] = {1,   2,   3,   4,   5,   6};
    double weights[] = {  .2,   .1,  .4,  .25, .05};
    std::piecewise_constant_distribution<> dist(std::begin(interval),
                                                std::end(interval),
                                                std::begin(weights));
    // Choose generator
    std::mt19937 gen;  // seed as wanted
    // Demonstrate by pouring into avg[rand-1]
    const unsigned N = 1000000;
    double avg[sizeof(weights) / sizeof(weights[0])] = {0};
    for (unsigned i = 0; i < N; ++i)
        avg[static_cast<unsigned>(dist(gen)) - 1]++;
    // Comute averages
    for (double* i = std::begin(avg); i < std::end(avg); ++i)
        *i /= N;
    // Display
    for (unsigned i = 1; i <= sizeof(avg)/sizeof(avg[0]); ++i)
        std::cout << "avg[" << i << "] = " << avg[i-1] << '\n';
}

Which for me outputs:

avg[1] = 0.199779
avg[2] = 0.100002
avg[3] = 0.400111
avg[4] = 0.250257
avg[5] = 0.049851

Answer 6

What you are describing is the implementation of a random number generator that draws from a particular probability distribution. For example, drawing numbers from a Gaussian distribution should draw random numbers such that the probability of a particular draw, x is proportional to
_{(source: wikimedia.org)}
.

In general, the approach is to draw from a uniform random distribution and then pick the value of the desired distribution's cumulative distribution function (CDF) at that drawn location. In the case of a Normal Gaussian, draw a random number, x from a uniform distribution (this is what standard random number generators should give) and then choose as the random, Gaussian distributed value. For your case, the CDF you describe is a piece-wise continuous stair-step function which could be implemented using any of the many (correct) answers you have already received.

Of course, this is all trivia. What you should be doing is using a library that already handles this for you. Statistics and random number generation are not trivial and there's no need to re-invent the wheel. See Neil's answer (and check out the Boost random number library).